Sujet : Re: how
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : sci.mathDate : 10. Apr 2024, 00:06:05
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <uv4hkt$jm4k$1@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Mozilla Thunderbird
On 4/9/24 8:16 AM, WM wrote:
Le 09/04/2024 à 01:22, Richard Damon a écrit :
On 4/8/24 9:44 AM, WM wrote:
Le 07/04/2024 à 19:56, Richard Damon a écrit :
On 4/7/24 9:23 AM, WM wrote:
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So, With infinite sets, a proper subset CAN be the same size as its parent.
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Impossible.
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Nope, PROVEN.
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Proven impossble with my matrix,
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Nope, since you matrix doesn't follow the required form.
It does precilsely.
Nope, because ONE set is not TWO Sets.
And a piece of a set is not a Sub Set.
and we can build such a mapping between the set of natural Numbers (N) with the set of even Numbers (E).
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Only handwaving by "and so on"
In all cases there are infinitely many exceptions.
∀n ∈ ℕ_applied: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
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I didn't say "N_applied", I said N.
But what you can use belongs to ℕ_applied. Otherwise show a natural number that completes the bijection, i.e., which has not infinitely many pairings on front.
Nope, you can use ALL of the Natural Numbers.
If you can only "use" N_applied, your system doesn't actually HAVE the Natural numbers
You "Complete" the bijection by showing the infinite sets map one to one by the formula of the bijection.
No "Last" one.
That seems to be your problem, you just can't handle INFINITE sets by your logic, and blow it up trying.
Regards, WM
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